The frequency radiation emitted when the electron falls from $\mathrm{n}=4$ to $n=1$ in a hydrogen atom will…
The frequency radiation emitted when the electron falls from $\mathrm{n}=4$ to $n=1$ in a hydrogen atom will be (Given ionization energy of $\mathrm{H}=2.18 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$ and $h$ $\left.=6.625 \times 10^{-34} \mathrm{Js}\right)$ :
$1.54 \times 10^{15} \mathrm{~s}^{-1}$
$1.03 \times 10^{15} \mathrm{~s}^{-1}$
$3.08 \times 10^{15} \mathrm{~s}^{-1}$
$2.00 \times 10^{15} \mathrm{~s}^{-1}$
Solution
$\mathrm{E}=h \mathrm{v}$ or $\mathrm{v}=\frac{\mathrm{E}}{h}$
For $\mathrm{H}$ atom,
$\begin{aligned}
E & =\frac{-21.76 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{~atm}^{-1} \\
\Delta E & =-21.76 \times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)
\end{aligned}$
$\begin{aligned}
& =20.40 \times 10^{-19} \mathrm{~J} \mathrm{~atm}^{-1} \\
v & =\frac{20.40 \times 10^{-19}}{6.625 \times 10^{-34}} \\
& =3.079 \times 10^{15} \mathrm{~s}^{-1} .
\end{aligned}$